For `x gt 0, lim_(x to 0)((sin x)^(1//x) + (1/x)^(sin x))` is equal to:

Options

A. `-1`
B. 0
C. 2
D. 1

Answer

To solve the limit \( \lim_{x \to 0} \left( (\sin x)^{\frac{1}{x}} + \left(\frac{1}{x}\right)^{\sin x} \right) \), we will break it down into two parts: \( L_1 = \lim_{x \to 0} (\sin x)^{\frac{1}{x}} \) and \( L_2 = \lim_{x \to 0} \left(\frac{1}{x}\right)^{\sin x} \).

### Step 1: Evaluate \( L_1 = \lim_{x \to 0} (\sin x)^{\frac{1}{x}} \)

1. As \( x \to 0 \), \( \sin x \to 0 \). Therefore, we have the form \( 0^{\infty} \), which is indeterminate.
2. We can rewrite \( L_1 \) using the exponential function:
\[ L_1 = e^{\lim_{x \to 0} \frac{\log(\sin x)}{x}} \]

### Step 2: Evaluate \( \lim_{x \to 0} \frac{\log(\sin x)}{x} \)

1. Using the fact that \( \sin x \approx x \) as \( x \to 0 \), we can write:
\[ \log(\sin x) \approx \log(x) \]
2. Therefore, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\log(\sin x)}{x} \approx \lim_{x \to 0} \frac{\log(x)}{x} \]
3. This limit is of the form \( \frac{-\infty}{0} \), which is also indeterminate.

### Step 3: Apply L'Hôpital's Rule

1. Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx} \log(x) = \frac{1}{x}, \quad \text{Denominator: } \frac{d}{dx} x = 1 \]
2. Applying L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\log(x)}{x} = \lim_{x \to 0} \frac{1/x}{1} = \lim_{x \to 0} \frac{1}{x} = -\infty \]

### Step 4: Conclude \( L_1 \)

1. Since \( \lim_{x \to 0} \frac{\log(\sin x)}{x} = -\infty \), we have:
\[ L_1 = e^{-\infty} = 0 \]

### Step 5: Evaluate \( L_2 = \lim_{x \to 0} \left(\frac{1}{x}\right)^{\sin x} \)

1. Rewrite \( L_2 \):
\[ L_2 = e^{\lim_{x \to 0} \sin x \cdot \log\left(\frac{1}{x}\right)} = e^{\lim_{x \to 0} \sin x \cdot (-\log x)} \]
2. As \( x \to 0 \), \( \sin x \approx x \), so:
\[ \lim_{x \to 0} \sin x \cdot (-\log x) \approx \lim_{x \to 0} x \cdot (-\log x) \]

### Step 6: Evaluate \( \lim_{x \to 0} x \cdot (-\log x) \)

1. This limit is of the form \( 0 \cdot \infty \), which is indeterminate.
2. Rewrite it as:
\[ \lim_{x \to 0} \frac{-\log x}{1/x} \]
3. Apply L'Hôpital's Rule:
\[ \text{Numerator: } \frac{d}{dx} (-\log x) = -\frac{1}{x}, \quad \text{Denominator: } \frac{d}{dx} (1/x) = -\frac{1}{x^2} \]
4. Thus:
\[ \lim_{x \to 0} \frac{-\log x}{1/x} = \lim_{x \to 0} \frac{-\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0} x = 0 \]

### Step 7: Conclude \( L_2 \)

1. Since \( \lim_{x \to 0} x \cdot (-\log x) = 0 \), we have:
\[ L_2 = e^0 = 1 \]

### Step 8: Combine \( L_1 \) and \( L_2 \)

1. Now, we can combine the results:
\[ L = L_1 + L_2 = 0 + 1 = 1 \]

### Final Answer

Thus, the limit is:
\[ \lim_{x \to 0} \left( (\sin x)^{\frac{1}{x}} + \left(\frac{1}{x}\right)^{\sin x} \right) = 1 \]